Question

Triangle is formed by the lines $x+y=0,x-y=0and\ell x+my=1$. If $\ell andm$ follow the condition ${\ell }^2+m^2=1$; then the locus of its circumcentre is

• A. $(x^2-y^2{)}^2=x^2+y^2$
• B. $(x^2+y^2{)}^2=(x^2-y^2)$
• C. $(x^2+y^2)=4x^2{y}^2$
• D. $(x^2-y^2{)}^2(x^2+y^2)=1$

Answer 1

The correct option is A $(x^2-y^2{)}^2=x^2+y^2$

Given lines are $x+y=0.......\left(i\right)x-y=0......\left(ii\right)$

$lx+my=\mathrm{1....}\left(iii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, we get

$x=0,y=0$

Solving $\left(ii\right)$ and $\left(iii\right)$, we get

$x=\frac{1}{\ell +m}$, $y=\frac{1}{\ell +m}$

Solving $\left(i\right)$ and $\left(iii\right)$, we get

$x=\frac{1}{\ell -m}$, $y=\frac{1}{m-\ell }$

Centroid $x_1=\frac{\frac{1}{m+\ell }+\frac{1}{\ell -m}+0}{3}$
$=\frac{\ell -m+m+\ell }{3({\ell }^2-m^2)}=\frac{2\ell }{3({\ell }^2-m^2)}$
$y_1=\frac{\frac{1}{m+\ell }+\frac{1}{m-\ell }+0}{3}=\frac{2m}{3(m^2-{\ell }^2)}$
Co-ordinates of circumcentre
$=\left(\begin{array}{c}\frac{\ell }{{\ell }^2-m^2},\frac{m}{m^2-{\ell }^2}\end{array}\right)$
$h=\frac{\ell }{{\ell }^2-m^2}.....\left(1\right)and{\ell }^2+m^2=1andk=-\frac{m}{{\ell }^2-m^2}.....\left(2\right)$
Square & add (1) and (2)
$h^2+k^2=\frac{{\ell }^2+m^2}{({\ell }^2-m^2{)}^2}=\frac{1}{({\ell }^2-m^2{)}^2}$
$\frac{1}{({\ell }^2-m^2{)}^2}=(h^2-k^2{)}^2$
Locus is $x^2+y^2=(x^2-y^2{)}^2$.